NCERT Solutions
Class 10 - Mathematics - Chapter 5: ARITHMETIC PROGRESSIONS - Exercise 5.3

Question 2

Recall: For an AP with first term \(a\), common difference \(d\), and \(n\) terms,

\[a_n = a + (n - 1)d, \quad S_n = \frac{n}{2}(a + a_n)\]

(i) \(7 + 10\tfrac{1}{2} + 14 + \ldots + 84\)

First, write the terms in decimal or fraction form:

\[7,\ 10\tfrac{1}{2} = 10.5,\ 14,\ldots,\ 84\]

First term: \(a = 7\).

Second term: \(10.5\). So the common difference is

\[d = 10.5 - 7 = 3.5\]

We know the last term \(a_n = 84\). Use \(a_n = a + (n - 1)d\):

\[84 = 7 + (n - 1)\cdot 3.5\]

\[(n - 1)\cdot 3.5 = 77\]

\[n - 1 = \frac{77}{3.5} = 22 \Rightarrow n = 23\]

Now use the sum formula:

\[S_{23} = \frac{23}{2}(7 + 84) = \frac{23}{2} \times 91\]

\[S_{23} = \frac{23 \times 91}{2} = \frac{2093}{2} = 1046\tfrac{1}{2}\]

So, the sum is \(1046\tfrac{1}{2}\).

(ii) \(34 + 32 + 30 + \ldots + 10\)

Sequence: \(34, 32, 30, \ldots, 10\).

First term: \(a = 34\).

Common difference:

\[d = 32 - 34 = -2\]

Last term: \(a_n = 10\).

Use \(a_n = a + (n - 1)d\):

\[10 = 34 + (n - 1)(-2)\]

\[10 = 34 - 2(n - 1)\]

\[10 = 34 - 2n + 2 = 36 - 2n\]

\[2n = 36 - 10 = 26 \Rightarrow n = 13\]

Now find the sum:

\[S_{13} = \frac{13}{2}(34 + 10) = \frac{13}{2} \times 44\]

\[S_{13} = 13 \times 22 = 286\]

So, the sum is \(286\).

(iii) \(-5 + (-8) + (-11) + \ldots + (-230)\)

Sequence: \(-5, -8, -11, \ldots, -230\).

First term: \(a = -5\).

Common difference:

\[d = -8 - (-5) = -3\]

Last term: \(a_n = -230\).

Use \(a_n = a + (n - 1)d\):

\[-230 = -5 + (n - 1)(-3)\]

\[-230 = -5 - 3(n - 1)\]

\[-230 = -5 - 3n + 3 = -2 - 3n\]

\[-3n = -230 + 2 = -228 \Rightarrow n = 76\]

Now find the sum:

\[S_{76} = \frac{76}{2}(-5 + (-230)) = 38 \times (-235)\]

\[S_{76} = -8930\]

So, the sum is \(-8930\).